First selection test. a n = 3n + n 2 1. b n = 2( n 2 + n + n 2 n),

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1 First selection test Problem. Find the positive real numbers a, b, c which satisfy the inequalities 4(ab + bc + ca) a 2 + b 2 + c 2 3(a 3 + b 3 + c 3 ). Laurenţiu Panaitopol Problem 2. Consider the numbers and for all n =, 2,..., 49. a n = 3n + n 2 b n = 2( n 2 + n + n 2 n), Prove that there are integers A, B so that a b + a 2 b a n b n = A + B 2. Titu Andreescu Problem 3. Consider a circle of center O and V a point outside the circle. The tangents from V touch the circle at points T, T 2. Let T be a point on the small arc T T 2 of the circle. The tangent at T intersects the line V T in the point A and the lines T T and V T 2 intersect in the point B. Let M be the intersection point of the lines OM and AB. Prove that lines OM and AB are perpendicular. Mircea Fianu Problem 4. Consider a cube and let M, N be two of its vertices. Assign the number to these vertices and 0 to the other six vertices. We are allowed to select a vertex and to increase with a unit the numbers assigned to the 3 adjiacent vertices - call this a movement. Prove that there is a sequence of movements after which all the numbers assigned to the vertices of the cube became equal if and only if MN is not a diagonal of a face of the cube. Marius Ghergu, Dinu Şerbănescu

2 2 Second selection test Problem 5. Let ABC be an acute triangle and let D be a point on the side BC. Points E and F are the projections of the point D on the sides AB and AC, respectively. Lines BF and CE meet at point P. Prove that AD is the bisector line of the angle BAC if and only if lines AP and BC are perpendicular. Sever Moldoveanu Problem 6. An array 8 8 consists of 64 unit squares. Inside each square are written the numbers or so that in any 2 2 subarray the sum of the four numbers equals 2 or 2. Prove that there exist two rows of the array which are equal. Marius Ghergu Problem 7. Consider a triangle ABC with the side lenghts a, b, c so that a is the greatest. Prove that the triangle is rightangled if and only if ( a + b + a b)( a + c + a c) = (a + b + c) 2. Virgil Nicula Problem 8. Find all positive integers n for which there are distinct integer numbers a, a 2,..., a n such that n = a + a a n. a a 2 a n 2 Dinu Şerbănescu

3 3 Third selection test Problem 9. In a chess tournament each of the players have played with all the others two games, one time with the white pieces and then with the black pieces. In each game the winners gets one point and both players receive 0.5 points if the game ends with a draw. At the end of the tournament, all the players end with the same number of points. a) Prove that there are two players with the same number of draws. b) Prove that there are two players with the same number of losses when playing the white. Marius Ghergu Problem 0. Consider the triangle ABC with AB = AC and a variable point M on the line BC so that B is between M and C. Prove that the sum between the inradius of AMB and the exradius of AMC corresponding to the angle M is constant. Problem. Let p, q, r be primes and let n be a positive integer such that Prove that n =. p n + q n = r 2. Problem 2. Let a < b c < d be positive integers so that ad = bc and d a. Prove that a is a square. Virgil Nicula Laurenţiu Panaitopol Dinu Şerbănescu

4 4 Fourth selection test Problem 3. Let ABC be a triangle inscribed in the circle K and consider a point M on the arc BC that do not contain A. The tangents from M to the incircle of ABC intersect the circle K at the points N and P. Prove that if BAC = N M P, then triangles ABC and M N P are congruent. Problem 4. The real numbers a, a 2,..., a n satisfy the relation Prove that a k 0, for all k =, 2, a 2 + a a (a + a a 00 ) 2 = 0. Valentin Vornicu Dinu Şerbănescu Problem 5. A finite set of positive integers is called isolated if the sum of the elements in any proper subset is a number relatively prime with the sum of the elements of the isolated set. Find all nonprime integers n for which there exist positive integers a, b so that the set A = {(a + b) 2, (a + 2b) 2,..., (a + nb) 2 } is isolated. Gabriel Dospinescu Problem 6. A regular polygon with 000 sides has the vertices colored in red, yellow or blue. A move consists in choosing to adjiacent vertices colored differently and coloring them in the third color. Prove that there is a sequence of moves after which all the vertices of the polygon will have the same color. Marius Ghergu

5 5 Problem 7. Consider the triangular array Fifth selection test defined by the conditions: i) on the first two rows, each element starting with the third is the sum of the two preceding elements; ii) on the other rows each element is the sum of the two elements placed above of the same column. a) Prove that all the rows are defined according to condition i). b) Consider 4 consecutive rows and let a, b, c, d be the first element in each of these rows, respectively. Find d in terms of a, b and c. Dinu Şerbănescu Problem 8. Let M, N, P be the midpoints of the sides BC, CA, AB of the triangle ABC, respectively, and let G be the centroid of the triangle. Prove that if BMGP is cyclic and 2BN = 3AB, then triangle ABC is equilateral. Problem 9. Let A be a set of positive integers with the properties: i) if a A, then all positive divisors of a are elements of A; ii) if a, b A and < a < b, then + ab A. Prove that if the set A has at least 3 elements, then A = N. BMO shortlist 2004 Valentin Vornicu Problem 20. Consider a convex polygon with n 5 sides. Prove that there are at most n(2n 5) 3 triangles of area with the vertices among the vertices of the polygon. Andrei Neguţ

6 6 FORMAL SOLUTIONS Solution to problem. Using the Chebyshev inequality we derive hence a + b + c. On the other hand, therefore ab + bc + ca. As (a + b + c)(a 2 + b 2 + c 2 ) 3(a 3 + b 3 + c 3 ), 4(ab + bc + ca) a 2 + b 2 + c 2 ab + bc + ca, 3(ab + bc + ca) (a + b + c) 2, we obtain a + b + c, thus a + b + c =. Consequently, a + b + c = and 3(ab + bc + ca) = (a + b + c) 2, which imply a = b = c = 3. Solution to problem 2. The key idea is to observe that a n b n = 2 ( n 2 n + n + ) 2. As 2 n n + n, it follows that the sum is Solution to problem 3. The approach of the problem is to see no circles in the figure. Instead, recall that a quadrilateral ABCD is orthogonal if and only if () AB 2 + CD 2 = AD 2 + BC 2. Using succesively the Pytagoras theorem we have BA 2 BT 2 2 = BA 2 (BO 2 OT 2 2 ) = BA 2 (BO 2 OT 2 ) = BA 2 (BA 2 AT 2 ) = AT 2 = AO 2 OT 2 = OA 2 OT 2 2, so the conclusion follows from the relation (). Solution to problem 4. Color the 8 vertices of the cube in black or white so that the 4 vertices of the 2 regular tetrahedrons have the same color; notice that the 3 vertices adjiacent to a vertex have its opposite color. Therefore, each movement increase the sum of the numbers assingned to the vertices sharing the same color by 3. Consider the cases: ) MN is a diagonal of a face of the cube. Then M and N have the same color, say black.assume by contradiction that there is a sequence of movements after which the same number n is assigned to all the vertices. Let k and k 2 be the number of white, respectively black vertices that were selected to perform the movements.then 4n = 3k + 2 = 3k 2, a contradiction.

7 7 2) MN is a diagonal of the cube. Selecting the vertices M, then N, and performing these 2 movements, to all the vertices the number will be assigned, as needed. 3) MN is a side of the cube. The same outcome as in the previous case will occur after 2 movements when selecting the diagonally opposite vertices of M and N. This provides us with a full solution. Solution to problem 5. Let a, b, c, x.y be the lengths of the sides BC, CA, AB, BD, DC, respectively and let A be the foot of the altitude from A in the triangle ABC. Notice that x + y = a. Due to Ceva theorem the claim is equivalent to BD DC CF F A AE EB = AB AC = BD DC. As CF = y cos C, F A = b y cos C, BE = x cos B, AE = c x cos B, BA = c cos B and A C = b cos C, the equivalence rewrites cy(c x cos B) = bx(b y cos C) xb = cy. Indeed, we have c 2 y cxy a2 + c 2 b 2 = b 2 x bxy a2 + b 2 c 2 2ac 2ab a(c 2 y b 2 x) = xy(c 2 b 2 ) c 2 y(a x) = b 2 x(a y) c 2 y 2 = b 2 x 2 cy = bx, as claimed. Solution to problem 6. The main idea is to observe that two consecutive rows have exactly 4 equal elements, namely those lying on the columns, 3, 5, 7 or 2, 4, 6, 8. Moreover, on the other 4 columns the elements are different. Wlog, assume that rows and 2 are equal with respect to the columns, 3, 5, 7 and different on the column 2, 4, 6, 8; we call these rows odd equal. If rows 2 and 3 are also odd equal, then rows and 3 are equal, as needed. If not, then rows 2 and 3 are even equal. Now consider the rows 3 and 4; we are done if the rows are even equal, so assume that they are odd equal. Finnaly, if rows 4 and 5 are odd equal,then rows 3 and 5 are equal, and if rows 4 and 5 are even equal, then rows and 5 are equal. This concludes the proof. Solution to problem 7. Squaring both sides of the equality yields 2(a + a 2 b 2 )(a + a 2 c 2 ) = (a + b + c) 2. It is easy to observe that the equality holds if a 2 = b 2 + c 2. To prove the converse statement, assume that a 2 > b 2 + c 2. Then a 2 b 2 > c and a 2 c 2 > b, hence 2(a + a 2 b 2 )(a + a 2 c 2 ) > 2(a + b)(a + c) = 2a 2 + 2(ab + bc + ca) > a 2 + b 2 + c 2 + 2(ab + bc + ca) = (a + b + c) 2, false. The case a 2 b 2 + c 2 leads similarily to a contradiction and we are done.

8 8 Solution to problem 8. Rearrange the numbers a, a 2,..., a n in ascending order: b < b 2 <... < b n. Obviously, k b k, and substituting b k with k, the left-hand side term increases. Futhermore, by the Rearrangements inequality we infer that the maximum value of the left-hand side term is n + 2 n + + n. On the other side, the right-hand side term is greater than or equal to We have = (n + ) n 2 = n + 2 n + + n = n k= For n > 6 we prove by induction on n that n(n + ). 4 n k= n k n = + (n + ) n+ n 4 k, k=2 k=2 n k + k n+ k = (n + ) which implies that the given equality cannot hold. Indeed, for n = 7 we have 7 4 = = If the inequality holds for n > 7 then it is true for n +, as 4 n+. We are left with the cases when n = 2, 3, 4, 5, 6. Clearly, the case case n = 2 is impossible. For n = 3 we have the numbers a =, a 2 = 2 and a 3 = 3., so n = 3 is a solution. If n = 4, then a + a 2 + a 3 + a 4 = 2( a + 2 a a a 4 ) 2( ) < 3, so a + a 2 + a 3 + a 4 2. By inspection, all the cases: {a, a 2, a 3, a 4 } = {, 2, 3, 4}, {, 2, 3, 5}, {, 2, 4, 5} and {, 2, 3, 6} fail to satisfy the required relation. If n = 5, then a + a 2 + a 3 + a 4 + a 5 = 2( a + 2 a a a a 5 ) 2( ) < 7.4, so a + a 2 + a 3 + a 4 + a 5 7. We study the cases {a, a 2, a 3, a 4, a 5 } = {, 2, 3, 4, 5}, {, 2, 3, 4, 6} and {, 2, 3, 5, 6} with no succes (for an easy argument, observe that 5 must be a 5 and so on). Finally, for n = 6 we obtain similarly a + a a 6 22, thus {a, a 2, a 3, a 4, a 5, a 6 } can be {, 2, 3, 4, 5, 6} or {, 2, 3, 4, 5, 7}. The last case fails immediately because of 7, and the same outcame is for the first one. Therefore n = 3. k=2 k.

9 9 Solution to problem 9. Let n be the number of players in the tournament. The total numbers of matches is n(n ), hence each player end up with n points. a) Assume by contradiction that each player has a different number of draws. As a draw gives 0.5 points, it follows that each player has an odd number of draws. Since the possible cases are: 0, 2, 4,..., 2(n ), we infer that each of these numbers is assigned to each of the players. Consider A the player with 0 draws and B the player with 2(n ) draws. Each player has played 2(n ) matches, hence B obtained a draw in each match played. The match A-B thus ended with a draw, a contradiction, since A has no draws. b) Suppose the contrary. Then each of the n players has 0,,..., n losses when playing the white. Let X and Y be the players with 0 and n losses, respectively. The player Y has no points when playing the white and n points, so he won all the matches with the black pieces. This implies that the match X-Y is won by Y, so is lost by X, a contradiction, since X has 0 losses with the white pieces. Solution to problem 0. The idea is to prove that the sum of the radii is equal to the altitude h from A of the triangle ABC; a hint is to think that line MA is parallel to BC. The Stewart relation gives AM 2 BC + AC 2 MB = AB 2 MC + MB MC BC, so AM 2 BC = AB 2 (MC MB) + MB MC BC, hence AM 2 = AB 2 + MB MC. Let r be the inradius of triangle AMB and R the exradius of triangle AMC corresponding to the angle M. Since and then and Thus so the claim holds. r = R = 2 areaamb AM + MB + AB, 2 areaamc MA + MB AB, r h = MB AM + MB + AB, R h = MC AM + MC AB. r + R = h MB(MA + MB AB) + MC(MA + MB + AB) = (MA + MB + AB)(MA + MB AB) MB(MA + MB AB) = (MA + MB + AB)(MA AB) MB MC + MB(MA AB) = MA 2 AB 2 + MB(MA AB) MA 2 = AB 2 + MB MC,

10 0 Solution to problem. Clearly one of the primes p, q or r is equal to 2. If r = 2 then p n + q n = 4, false, so assume that p > q = 2. Consider the case when n > is odd; we have (p + 2)(p n 2p n p n n ) = r 2. Notice that p n 2p n p n n = 2 n + (p 2)(p n 2 + p n 4 + ) > and p + 2 > hence both factors are equal to r. This rewrites as p n + 2 n = (p + 2) 2 = p 2 + 4p + 4, which is false for n 3. Consider the case when n > is even and let n = 2m. It follows that p m = a 2 b 2, 2 m = 2ab and r = a 2 + b 2, for some integers a, b with (a, b) =. Therefore, a and b are powers of 2, so b = and a = 2 m. This implies p m = 4 m < 4 m, so p must be equal to 3. The equality 3 m = 4 m fails for m = and also for m 2, as 4 m > 3 m +, by induction. Consequently n = - take for example p = 23, q = 2 and r = 5. Solution to problem 2. Consider the integers 0 < m n < p so that b = a+m, c = a+n and d = a+p. Then a(a+p) = (a+m)(a+n) and a + p a a. As p = m + n + mn a is an integer, then a mn and p m + n +. On the other hand, + 2 a n + m + a m+n 2 mn, hence a mn. Consequently, a = mn and m = n, so a is a square. Solution to problem 3. Let Q be the intersection point of the line segments AB and MP. The tangents from A and M to the incirle are equal (as they are r cot A 2 ). Moreover, the tangents from Q to the incircle are equal, so AQ = MQ. This implies QMA = QAM, so the arcs AP and BM are equal. In the trapezoid AP BM, the diagonals AB and MP are equal, and likewise AC = MN. This concludes the proof. Solution to problem 4. Assume by contradiction that a k > 0, for some k. Wlog, let k =. Then a 2 > 00 and a 2 2+a a 00 +s 2 <, where s = a + a a 00. On the other hand, the Cauchy-Schwarz inequality yields a contradiction. Solution to problem 5. a 2 = (s a 2 a 3... a 00 ) 2 00(a a a s 2 ) < 00, The sum of the elements of the set A is S = na 2 + n(n + )b + n(n+)(2n+) 6 b 2. Assume that n has a prime divisor p > 3. Then p S and p (a + b) 2 + (a + 2b) (a + pb) 2 = pa 2 + p(p + )b + p(p+)(2p+) 6 b 2, a contradiction. It remains n = 2 k 3 l, for some integers k, l. Suppose that k >. Then 2 S and so all elements of A must be odd. Taking the subset given by any pair we reach the contradiction. Finally, suppose that l >, so 3 S. If one of the numbers a + b, a + 2b, a + 3b is divisible by 3, then we have a contradiction; if not, then 3 b. Then 3 (a + b) 2 + (a + 2b) 2 + (a + 3b) 2 = 3a 2 + 2ab + 4b 2, again a contradiction. We are left with n = 6, which satisfies the claim: the set A = {4, 9, 6, 25, 36, 49} is isolated, because the sum of its elements is a prime number (39).

11 Solution to problem 6. Let A, A 2,... A 000 be the vertices of the polygon. We start with two lemmas. Lemma. Three of four consecutive vertices have the same color. Then after a sequence of moves all vertices will have the color of the fourth vertex. Proof. Let the colors be 0, and 2. We have two cases: a) b) Lemma 2. Any 4 consecutive vertices will turn after several moves in the same color. Proof. Form two pairs of consecutive vertices and change them in the same color - if they do not already have it. Then follow the sequence By the second lemma, after several moves the vertices A, A 2, A 3, A 4 will have the same color, say red. Likewise, A 5, A 6, A 7, A 8 will have the same color. Consider now the vertices A 4, A 5, A 6, A 7 ; the first is red and the other three have the same color. By the first lemma they all will turn red - of course, we do nothing if they were already red. We move on with this procedure until A, A 2,... A 997 turn red (note that 997 = , so this requires 332 steps). Now consider the vertices A 998, A 999, A 000, A ; by the second lemma they all will share the same color. If this is red, we are done. If not, say that they are blue, and taking the vertices A 997, A 998, A 999, A 000 we obtain - using the first lemma - all vertices to be red, except for A, which is blue. Now A, A 2, A 3, A 4 turn blue, then A 5, A 6, A 7, A 8 and so on. This time, after 333 steps, all the 000 vertices (000=+3 333) will be colored in blue. Comment. Substituting colors with digits, notice that all moves: 0 22, 02 and 2 00 preserve the sum (mod 3). This means that the final color is unique and, of course, is given by the sum of the digits assigned to the vertices of the initial configuration. Solution to problem 7. a) In the array below a b c d e f g h i where i is an element of the third row, observe that i = c + f = (a + b) + (d + e) = (a + d) + (b + c) = g + h. The same argument holds for all the other rows, by induction. b) We prove that d = 2b + 2c a. Indeed, from the array a x y z b t u c v d we derive d = u+v = b+t+(u+z) = 2(b+t)+(x+y)= 2b+2(t+y)+x y = 2b+2c+x (x+a) = 2b+2c a.

12 2 Solution to problem 8. By the Power of a point theorem we have AG AM = AP AB, so 4MA 2 = 3AB 2 and thus AM = 3 2 AB = BN. Then AG = GB, so the median GP is also an altitude in the triangle AGB. This implies BP G = 90, and since BMGP is cyclic, GMA = 90. It follows that BC = CA and AB = AC, so the triangle is equilateral. Solution to problem 9. It is obvious that A, since is a divisor of any integer. Consider a, b two elements of A with < a < b. Since at least one of a, b or + ab is even, then 2 is an element of A. We induct on n 6 to prove that n A. Assume that k A for all k =, 2,..., n. If n is odd, then n = 2p+ with < 2 < p A, hence n A. If n is even, then n = 2p. As above, 2p and 2p+ are elements ofa and consequently + (2p )(2p + ) = 4p 2 A. The first property implies n = 2p A, as needed. To complete the proof, we need to show that 3, 4, 5 A. For this, consider a > 2 an element of A. Then + 2 a A, + 2( + 2a) = 3 + 4a A and + ( + 2a)(3 + 4a) = 4 + 0a + 8a 2 A. If a is even, then a + 8a 2 and so 4 A. If a is odd, then choose a to be 4 + 0a + 8a 2 and again 4 A. Next, as < 2 < 4 A we have = 9 A and so 3 A. Finally, 7 = A, 5 = A, hence 5 A and we are done. Solution to problem 20. Let A, A 2,... A n be the vertices of the polygon. We start with the following Lemma:Each segment A i A j belongs to at most 2 triangle of area located on the same side of the line A i A j. Proof of the lemma. Indeed, suppose that on the same side of the line A i A j exist the vertices A m, A n, A p so that the triangles A i A j A m, A i A j A n and A i A j A p have the area. Then the points A m, A n, A p will be at the same distance to the line A i A j, hence colinear. This is a contradiction, since the polygon is convex. Consider first the n sides of the polygon. Each of them can form at most 2 triangles of area, as all the vertices lie on the same side, hence we have by now at most 2n such triangles. Consider now the n diagonals A i A i+2 - with the cyclic notations: A n+j = A j. Each of them can form at most 3 triangles of area, one with A i+ and two with the vertices lying on the other side. Thus we have at most 5n = 2n + 3n triangles. Finally, consider the other diagonals of the polygon. They are n(n 5) 2, and each of them can form at most 4 triangles. The final counting is 5n + 4 n(n 5) 2 = n(2n 5), except that we have counted each triangle three times, one time for each side. Therefore, there are at most n(2n 5) 2 triangles, as claimed.